Question

Calculate the degree of ionization and \[\text{ pH }\] of \[\text{ 0}\text{.05 M }\] the solution of a weak base having the ionization constant \[\text{ }\left( {{\text{K}}_{\text{C}}} \right)\text{ }\] is\[\text{ }1.77\text{ }\times \text{ }{{10}^{-5}}\text{ }\] . Also, calculate the ionization constant of the conjugate acid of this base.

Answer 1

Hint: According to the Arrhenius concept, the base is the substance that dissociates into hydroxide ions. Here, the reaction of the dissociation of the base is given as:
$\text{ BOH }\rightleftharpoons \text{ }{{\text{B}}^{\text{+}}}\text{ + O}{{\text{H}}^{-}}$
The equilibrium constant or the \[\text{ }{{\text{K}}_{\text{b}}}\text{ }\]is related to the concentration of the solution and the degree of dissociation. It is the extent to which the base dissociates into the solution. The sum of $\text{pOH}$ and $\text{pH}$ is equal to the 14.

Complete step by step answer:
Let’s consider a weak monobasic base $\text{ BOH }$, its dissociation by the Arrhenius acid-base concept may be represented by the following equation:
$\text{ BOH }\rightleftharpoons \text{ }{{\text{B}}^{\text{+}}}\text{ + O}{{\text{H}}^{-}}$
The equilibrium constant or the dissociation constant \[\text{ }{{\text{K}}_{\text{b}}}\text{ }\]of the weak base is represented as:
\[\text{ }{{\text{K}}_{\text{b}}}\text{=}\dfrac{\left[ {{\text{B}}^{\text{+}}} \right]\left[ \text{O}{{\text{H}}^{-}} \right]}{\left[ \text{BOH} \right]}\text{ }\]
If the initial concentration of the weak base is ‘c’ moles per litre and if $\text{ }\alpha $ is the degree of dissociation, then the weak monobasic base dissociates as follows:
\[\begin{matrix}
   {} & \text{BOH} & \rightleftharpoons & {{\text{B}}^{+}} & + & \text{O}{{\text{H}}^{-}} \\
   \text{Initial conc}\text{.} & c & {} & 0 & {} & 0 \\
   \text{Final conc}\text{.} & c(1-\alpha ) & {} & c\alpha & {} & c\alpha \\
\end{matrix}\]
The dissociation constant $\text{ }{{\text{K}}_{\text{b}}}\text{ }$of the weak base is given as:
$\text{ }{{\text{K}}_{\text{b}}}\text{ = }\dfrac{c\alpha \text{ }\times \text{ }c\alpha \text{ }}{c(1-\alpha )}=\text{ }\dfrac{c{{\alpha }^{2}}}{(1-\alpha )}$
Since, $\alpha \ll 1$ , the dissociation constant of a weak base is rewritten as :
$\text{ }{{\text{K}}_{\text{b}}}\text{ = }c{{\alpha }^{2}}$
Let’s rearrange the equation concerning the dissociation constant we get,
\[\text{ }\alpha =\sqrt{\dfrac{{{\text{K}}_{\text{b}}}}{c}}\]
We have given the following data:
The dissociation constant of a weak base,${{\text{K}}_{\text{b}}}\text{ = }1.77\text{ }\times \text{ }{{10}^{-5}}\text{ }$
The concentration of weak base solution,$\text{ 0}\text{.05 M }$
To find:
a) Degree of ionisation
b) \[\text{ pH }\]Of solution
c) Ionisation constant of the conjugate acid
a) Lets, calculate the value of the degree of dissociation,
\[\text{ }\alpha =\sqrt{\dfrac{{{\text{K}}_{\text{b}}}}{c}}\text{ = }\sqrt{\dfrac{1.77\times {{10}^{-5}}}{0.05}}\text{ = 0}\text{.0188}\]
Therefore, the degree of dissociation $(\alpha )$ is \[\text{0}\text{.0188}\] or \[\text{ 1}\text{.88}\times \text{1}{{\text{0}}^{\text{-2}}}\] .
b) Now, we are interested to find out \[\text{ pH }\]about the solution.
The concentration of hydroxide ions is given as:
$\left[ \text{O}{{\text{H}}^{-1}} \right]=\text{ c}\alpha \text{ }$
Let’s calculate the concentration of hydroxide ions.
$\left[ \text{O}{{\text{H}}^{-1}} \right]=\text{ c}\alpha \text{ = }\left( 0.05\text{ }\times \text{ 1}\text{.8}\times \text{1}{{\text{0}}^{\text{-2}}} \right)\text{ = 9}\text{.0}\times \text{1}{{\text{0}}^{\text{-4}}}$
We know that,
$\text{pOH = }-\log \left[ \text{O}{{\text{H}}^{-}} \right]$
Therefore, the $\text{pOH}$ is
$-\log \left[ 9.0\times {{10}^{-4}} \right]=\text{ 3}\text{.04}$
The sum of $\text{pOH}$ and $\text{pH}$ is equal to the 14. Therefore, the $\text{pH}$ of the solution is,
$\begin{align}
  & 14=\text{pH + pOH} \\
 & 14=\text{pH + 3}\text{.04} \\
 & \text{pH}=\text{14 }-\text{ 3}\text{.04} \\
 & \therefore \text{pH = 10}\text{.95} \\
\end{align}$
So, $\text{pH}$ of the weak base is$\text{10}\text{.95}$.
c) Let's calculate the ionisation constant of the conjugate acid of a weak base. The product of the dissociation constant of a base and its conjugate acid is always equal to the dissociation constant of water ${{\text{K}}_{\text{w}}}$ . Thus , on applying this we can calculate the ionisation constant of conjugate acid ${{\text{K}}_{\text{a}}}$.
\[\begin{align}
  & {{\text{K}}_{\text{w}}}\text{ = }{{\text{K}}_{\text{base}}}\text{ }\!\!\times\!\!\text{ }{{\text{K}}_{\text{conjugate acid}}} \\
 & \Rightarrow {{\text{K}}_{\text{conjugate acid}}}\text{ = }\dfrac{{{\text{K}}_{\text{w}}}\text{ }}{{{\text{K}}_{\text{base}}}} \\
\end{align}\]
Let’s substitute the values,
\[{{\text{K}}_{\text{conjugate acid}}}\text{ = }\dfrac{{{\text{K}}_{\text{w}}}\text{ }}{{{\text{K}}_{\text{base}}}}\text{ = }\dfrac{{{10}^{-14}}}{1.77\times {{10}^{-5}}}\text{ = 5}\text{.65 }\times {{10}^{-10}}\]
Thus, the ionisation constant of the conjugate acid is equal to the\[\text{5}\text{.65 }\times {{10}^{-10}}\].
Hence,
a) The degree of dissociation $\text{ }(\alpha)\text{ }$ of a weak base is\[\text{ 1}\text{.88}\times \text{1}{{\text{0}}^{\text{-2}}}\].
b) The \[\text{ pH }\] of the weak base solution is equal to$\text{10}\text{.95}$.
c) The ionisation constant of conjugate acid \[{{\text{K}}_{\text{conjugate acid}}}\] or weak base is equal to\[\text{5}\text{.65 }\times {{10}^{-10}}\].

Note: According to the Lowry-Bronsted acid-base theory, the base dissociates into the solution and produces its corresponding conjugate acid. The base loses the hydroxide ion, the reaction is as shown below:
$\begin{matrix}
   \text{BOH} & \text{+} & {{\text{H}}^{\text{+}}} & \rightleftharpoons & {{\text{B}}^{\text{+}}} & \text{+} & {{\text{H}}_{\text{2}}}\text{O} \\
   \text{(Base)} & {} & \text{(Acid)} & {} & \text{(Conjugate acid)} & {} & \text{(Conjugate base)} \\
\end{matrix}$
Thus, the conjugate acid concentration can be obtained from the equilibrium constant.